[CP5670]

you can never have too much math...

[Styxx]

Originally posted by CP5670
you can never have too much math...

Yes, you can. The only things you can never have too much are beautiful women, and Spacecrack™.

Have you read the book "Flight of the Dragonfly"? You'd like the aliens on it...

[Red5]

Originally posted by CP5670
Some of those are the same as the Latin letters, which makes it a bit easier.

A B G D E Z H Q I K L M N O P R S T U F C X Y W

I don't like them because they are used by mathematicians simply to look smart.

And then there are these two things that aren't even Greek letters:
Ã(x) (weierstrass P function symbol - just look at this doodle thing...what the heck?!)
¶z/¶x (partial derivative symbol - what is this, a mirrored 6?)

Let me try this two-part integral sign (another weird symbol); see how it looks...
ó
õ
not bad...

thats the greek alphabet guys not latin...

[Shrike]

Originally posted by CP5670
I do that all the time! Here's an interesting problem, for example:

But all that has no real relevance to FS..... whereas my calculation does.

[Styxx]

Originally posted by Shrike
But all that has no real relevance to FS..... whereas my calculation does.

And that's why we forgive you.

[CP5670]

But all that has no real relevance to FS..... whereas my calculation does.

um, um...the expression in the first problem ( ³Ö(x²-a²) ) could be an energy-time function for a reactor in a FS fighter, and with the appropriate values for a, one could determine the work output with that integral, and if the mass is given, the top speed of the fighter could also be calculated!

As for the other problem, think subspace.

thats the greek alphabet guys not latin...

yeah, as I said, many of the capital Greek letters are the same as Latin ones.

[Shrike]

Originally posted by CP5670
um, um...the expression in the first problem ( ³Ö(x²-a²) ) could be an energy-time function for a reactor in a FS fighter, and with the appropriate values for a, one could determine the work output with that integral, and if the mass is given, the top speed of the fighter could also be calculated!

As for the other problem, think subspace.

I don't see them actually being used for that, though.

[wEvil]

Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?

And you're forgetting stress factors from the ships' spaceframe.  Chances are the beam would stress it quite badly, if only from thermal expansion.

[Styxx]

Originally posted by wEvil
Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?

And you're forgetting stress factors from the ships' spaceframe.  Chances are the beam would stress it quite badly, if only from thermal expansion.

For the whole beam, I suppose.

And he's getting a low-end estimate, considering the simplest elements possible only. The fact that he considered the hull pure molybdenum indicates that...

[Shrike]

Originally posted by wEvil
Is the 3.7 MT figure for the entire beam duration or just a second (for instance) of the strike?

And you're forgetting stress factors from the ships' spaceframe.  Chances are the beam would stress it quite badly, if only from thermal expansion.

Basically what Styxx said.  I had a post, but I lost it.

Just one comment though, the energy required to vaporize a large chunk of armor is probably far higher than the energy that would go into heat-stressing the frame.

[Zeronet]

Originally posted by EdrickV


Download GetRight. And you could try Wings3D. The run time engine thing it needs is about 23 megs, but the Windows install file for the program itself is about 913k. (That said, I ought to finish my download of GMAX so I can really compare it to Wings3D.)

http://www.GetRight.com

I have and use DAP.

[CP5670]

Wait, I just checked something; 3.7MT isn't consistent with some of the other values. The FS bible and tech entry says that the original GTM-N1 Harbinger had a 5000MT (5GT) payload.

[Styxx]

Exactly. That's why we have to assume that the armour on the destroyers is much stronger than simple molybdenum...

[Sandwich]

BUMP!

Hey, CP - here's a pic of one of my favorite shirts - mean anything to ya?

[CP5670]

LOL, actually the math makes sense to me, but it appears to be a physics formula of some sort. (it's mostly MV calculus, but the variables are the sort commonly seen in mechanics) The F is probably net force, V is linear velocity, w is rotational velocity, R is radius, x is linear displacement and t is obviously time, but I can't really tell what the rest could mean. Notice the betas, rhos and omegas though. Actually, now that I look at it, it looks like a variational optimization problem of some sort with those compounded derivatives and the contour integral.

[Sandwich]

Originally posted by CP5670
LOL, actually the math makes sense to me, but it appears to be a physics formula of some sort. (it's mostly MV calculus, but the variables are the sort commonly seen in mechanics) The F is probably net force, V is linear velocity, w is rotational velocity, R is radius, x is linear displacement and t is obviously time, but I can't really tell what the rest could mean. Notice the betas, rhos and omegas though. Actually, now that I look at it, it looks like a variational optimization problem of some sort with those compounded derivatives and the contour integral.

*Reads CP's post*

Drat - Babelfish doesn't do Calculus-to-English.

I heard from someone that the appearance of 3 of those "f"s in a row means that the whole thing is nonsense...?

[CP5670]

LOL, you mean those ò things? That's just the integral sign, so three of them would be a triple integral.

The one thing that is a little annoying is that they have used all the different kinds of notations for derivatives in the same problem: they have the letter subscripts, the ¶f/¶x's, and the time derivatives denoted by the letters with the dots on top. Historically, calculus notation has been something of a farce, with everyone using their own system, and all of them lacking in some way.

[Carl]

Originally posted by Shrike
Well, that's a low-end estimate... a loose value of about 1 MT/s.

Now, to make that jive with the low gigaton levels of Harbingers, we have to postulate that FS armor is significantly stronger than normal molybdenum.  The term 'Collapsed Molybdenum' should give us a good hint.  If that refers to collapsed electron shells, it could theoretically take extreme amounts of energy before melting/vaporising.

that could also explain why they are so low.

[wEvil]

I thought molybdenum had a low shearing value, though?

Adding a kind of reflective coating to the outer hull could possibly deflect much of the energy from beams etc.

Bombs..

By all means, a low-gigaton bomb would completely annihalate an orion.  The FS bible and V on this count are totally wrong, unfortunately (antimatter less powerful than 1fusion+three salted fission bombs? bah!)

[HotSnoJ]

Originally posted by Zeronet
I'd be modelling but after downloading 30 megs of GMAX only for its zip to be corrupt, i dunno if i can take the chance again.

Use Blender for heavan's sake!

BTW It does have a mirror. I'll write a tutorial soon for it and others.

To mirror just press 'S' (size) and then 'X' or 'Y'. Sorry 'Z' doesn't work. Then press 'Ctrl' to do even sizing and get it back you size 1.00 on all sides.